1

Power transmission Components used to transmit power: gears, belt, clutch and brakes.

Gear Objective: Student must be able to force analysis, stress analysis using basic formula (Lewis) and AGMA (bending stress and surface stress) Type of gear: Spur gear only

Types of Gear a) Spur Gear

b) Helical gear

Teeth is parallel to axis of rotation Can transmit power from one shaft to another

parallel shaft

Teeth is inclined to the axis of rotation Smoother than spur Develop thrust load (helix angle) Can transmit power from one shaft to a

parallel and non-parallel shaft

2

c) Bevel gear

d) Worm gear

Terminologies

A pair of gears can represented as 2 circles d1 = Nm d2 = Nm

2 3

where: N : number of teeth m : module

note: mating gear must have same m

Teeth on conical surfaces Transmit power between two intersecting

shafts

Transmit power between two intersecting shafts

3

Module: is the ratio if diametral pitch and number of teeth m = d/N [mm] Face Width (F) : width of the tooth Addendum [a] : distance between top face of the tooth to pitch circle Dedendum [b] : distance between pitch diameter to bottom of the gear In the following : we only concentrates on full depth gear

4

Full depth tooth

When the offset occurs between pitch. not full-depth tooth, which is called stub Undercut: resulted from number of tooth is less than the minimum number of tooth suggested. Results: higher stresses at the root of the tooth. (Refer to section 13-7 interference)

Backlash : gap between mating tooth. The gap can be used for lubrication

5

Contact Ratio: the number of tooth in contact during meshing. Roughly spur gear (1.4 to 1.8 )

Pinion and Gear : pinion is driver and gear is driven

Gear Parameters Metric Module m = d/N

Imperial Diametral Pitch P= N/d (inverse to module)

One pair of gear must have the same module

Pressure Angle: 200, 22.5o, 250

Table 13-1: relationship between addendum (a) and dedundum (b) pp

696

Table 13-2: Available pitch diametral and module.

6

Gear Train

2 3

V2 = V3 Known that

itmin/revolution:n:where

60

dnV

3322

3322

ndnd

ndnd

Equation 1

For a pair of gear m2 = m3

3

3

2

2

N

d

N

d Equation 2

From Eq 1 and 2

2

3

3

2

3

2

n

n

N

N

d

d

Significance: d increases N increase d increases n reduces to reduce rpm requires small

pinion and larger gear and vice versa.

7

Gear Train (continued)

2 3 4

V2 = V3 and V3 = V4

Therefore V2 = V3 = V4

From previous formula:

3322

2

3

3

2

nNnN

n

n

N

N

Gear 2 and 3

2

3

2

3n

N

Nn … (1)

Gear 3 and 4

3

4

3

4n

N

Nn …(2)

Eq (1) in eq (2)

2

3

2

4

3

4n

N

N

N

Nn

FLn

numberstoothdrivenofproduct

numberstoothdrivingofproductn

where: nL : rotational speed of last gear (output) nF : rotational speed of first gear (input)

Train value numberstoothdrivenofproduct

numberstoothdrivingofproducte

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Planetary Gear

Gear 2: Sun gear Gear 3: Arm Gear 4: Planet Gear Gear 5: Ring Gear

Assumption Arm Fixed:

2

3

4 5

Train value 654

542

NNN

NNNe

4

2

3

5

9

3 MAGIC FORMULAE FOR FBD ANALYSIS ON GEAR

Torque ]Nm[W2

dT

t

Wt: tangential force

Speed ]s/m[60

dnV

D: pitch diameter in [m] n: rotational speed [rpm]

Power ]watts[VWH

t

10

Force Analysis (Free Body Diagram)

Input rpm direction: cw

To transfer power, T must exist. When the pinion rotates, tooth from gear against the movement direction Wt32 must against the direction of rotation

Wt32 = H / V T2 = Wt32. d2/2

Due to pressure angle, Wr32 (radial force) is generated

Wr32 = Wt32 tan

11

On Gear 3,

Wt23 and Wr23 must in the opposite direction. To be statically analytical, T3 is against Wt23

T3 = Wt23. d3/2 Note: Wt23 can be calculated using Wt32= H/ V, please remember that all the parameters must be based on gear 3.

Discuss example 13-7

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Example You are responsible to design a gear system for speed reducer. The speed reducer is a two stage reduction which each pinion has 18T (Gear 2 and 4 in Figure 1). One of the constraints is that the maximum allowable reduction is 10 at each stage. Based on this, answer the following questions. a. Suggest the two possible number of teeth for Gear 3 and 5 if the speed has to be reduced by 24 times. Note: if Gear 3 has X teeth and Gear 5 has Y teeth and vice versa, the answer is considered as one) Number of Tooth Gear 3 Gear 5

Combination 1 ________ ________ Combination 2 ________ ________

Assume gear 3 and 5 have 72 and 90 teeth respectively and comprises of m = 4 and 200 pressure angle gears. The motor is 8kW at 1000 rpm clockwise. b. Calculate the rpm of the output shaft. c. Draw and calculate all the resultant forces on all of the gears. d. Based on the above calculation, discuss the relationship

between torque and gear ratio.

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Syaf Keluaran

2

3

4

5

18T 18T

Figure 1

13

Example The figure below shows a dual output power transmission system. A 8kW motor with 1000 rpm in clockwise direction is attached to shaft 1 at A. 40% of the power is delivered to shaft 2 using gear 2 and 4 and the remaining 60% of the power is delivered to shaft 3 via gear 3 and 5. All the gear module is 2 mm (m = 2mm) with pressure angle of 200. Based on this information, answer the following questions

a) Draw the FBD for every gear and also calculate all the forces and torques if the speed of both output shafts have been reduced by 3.

b) Calculate N5 if torque on shaft 2 and 3 is equal and N4 is 72.

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Failure Types Bending: resulted from bending stress. Wt act on the tooth Lewis formula and AGMA Pitting: resulted from surface stress Repetition of high contact stresses Scoring: resulted from insufficiency of lubrication

Bending Stresses

Basic Formula: FmY

W t

*take note that there are two formulae in pair in the textbook (imperial and metric)

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Assumption Basic Formula

Cantilever Beam Problem

Ft is carried by one tooth only.

However, dynamic effects are present when a pair of gear at moderate and high speed. ( KV’)

FmY

WK t'

V … eq 14-1 pp738

Y: Lewis form factor Table 14-2 pp 718. (interpolation if it is required)

Dynamic Effect

05.3

V05.3K '

V

(cast iron, cast profile)

1.6

V1.6K '

V

(cut or milled profile)

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56.3

V56.3K '

V

(hobbed or shaped profile)

56.5

V56.5K '

V

(finishing process on gears:

shaved or ground profile)

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Fatigue Strength

Endurance Limit eedcbae SkkkkkS '

ute SS 5.0'

For steel

For cast iron

AGMA

18

Surface Factor ka

b

uta aSk

Machining process 265.051.4 uta Sk

19

Size Factor kb

mmdd

mmddk

e

eb

2545151.1

5179.224.1157.0

107.0

In the case of gear tooth d = de

Ftde 808.0

2

3Ymx (Eq 14-3)

xlt 4 (Eq b)

l = add + dedd (refer to Table 13-1) Stress Concentration Factor

21 fff KKK

Stress concentration factor due to load (Kf1) kf1 = 1.66 (one way bending) kf1 = 1 (two way bending)

1

1

1

f

fk

K

Stress concentration 2 (Kf2)

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Refer to Table A-15-6

r = 0.3m d = t D = infinite

Based on r/d and D/d = infinite, uses the largest D/d, get the value of Kt

Kf2= 1 + q (Kt – 1) Definition of one and two way bending One way bending: tooth subjected one direction rev.

Two way: tooth subjected to both direction rev (forward and reverse)

Fatigue failure theories

Goodman Line nS

K

S

K

e

af

ut

mf 1

Gerber Failure 1

2

e

af

ut

mf

S

nK

S

nK

Determine the mean and alternating stress One way bending

21

2

am

Two way bending

a

0m

Where FmY

WK t

V

Project

You have to set m, F, NP, NG The constraint

Min no of tooth for pinion: Eq 13-11 Max no of tooth for gear: Eq 13-12 Face width 2p < F < 5p Material : Figure 14-2, 14-3

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SURFACE DURABILITY Surface Stresses (compressive –ve)

2

1

tVPC

2r

1

1r

1

cosF

WKC

= pressure angle P = pinion G = gear

Cp = elastic coefficient

2

1

G

2

G

P

2

P

P

E

1

E

1

1C

= Poisson Ration (Table A-5 ) E = Modulus of Elasticity (Table A-5)

Radius of curvature of the tooth profile

2

sind2r

2

sind1r

G

p

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Example A 19-TOOTH 300 Bhn HOBBED STEEL SPUR GEAR PINION TRANSMITS 15 Kw AT A PINION SPEED OF 360 rev/min TO A 77 TOOTH OF THE SAME MATERIAL GEAR. THE FACE WIDTH IS 75 mm, = 200 AND m = 6mm. a) USING LEWIS FORMULA CALCULATE THE STRESSES DUE

TO BENDING AND THE CONTACT STRESSES? b) CALCULATE THE F.S OF THE BENDING STRESS AGAINST

ITS FATIGUE STRENGTH USING GERBER THEORY? c) CALCULATE THE F.S OF THE CONTACT STRESSES

AGAINST CONTACT STRESS ENDUCRANCE LIMIT (Sc)?

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Soln

a) Bending Stresses

Pinion

FmY

WK t

v

d = Nm = (19)(6) = 114 mm = 0.114 m

]s/m[60

)360)(114.0(

60

dnV

= 2.149 m/s

149.2

000,15

V

HWt = 6980 N

Hobbed 56.3

V56.3K '

V

= 1.402

Y = 0.314 (Table 14-2)

FmY

WK t

v = 69.8 MPa

GEAR

Y)6)(75(

)6980)(412.1(

FmY

WK t

v

Y interpolation (77 tooth)Y = 0.436

Y)6)(75(

)6980)(412.1(

FmY

WK t

v = 50.23 MPa

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b) Surface Stresses

2

1

tVPC

2r

1

1r

1

cosF

WKC

Kv = 1.40 Wt = 6980 N F = 75mm dp = Nm = (19)(6) = 114 mm dg = Nm = (77)(6) = 462 mm

2

sind1r

p = 19.5 mm

2

sind2r G = 79 mm

2

1

G

2

G

P

2

P

P

E

1

E

1

1C

Carbon Steel: V = 0.292 E = 207 GPa = 207 x 103 MPa Cp = 189.780 MPa

2

1

2

1

tVPC

79

1

5.19

1

)20cos()75(

)6980)(402.1()780,189(

2r

1

1r

1

cosF

WKC

= -565.5 MPa

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Safety factor

MPa)10H4.0(89.6S

Sn

Bc

c

c

Sc = 757.9 n = 1.15

27

Limits

S.F

Skkkkk

S.F

S '

eedcbae

all

It is advisable to have F.S > 3

Contact Stress

2

1

21

11

cos

rrF

WKC

t

v

pc

… Eq 14-14 pp 732

Negative: because it is always in compression Cp …. Eq 14.13 E : Modulus of elasticity Table A-5

u : Poisson’s Ratio Table A-5

r1 , r2 : Eq 14-12

28

Example You are responsible to design a gear system for speed reducer. The speed reducer is a two stage reduction which each pinion has 18T (Gear 2 and 4 in Figure 1). One of the constraints is that the maximum allowable reduction is 10 at each stage. Based on this, answer the following questions. a. Suggest the two possible number of teeth for Gear 3 and 5 if the speed has to be reduced by 24 times. Note: if Gear 3 has X teeth and Gear 5 has Y teeth and vice versa, the answer is considered as one) Number of Tooth Gear 3 Gear 5

Combination 1 ________ ________ Combination 2 ________ ________

Assume gear 3 and 5 have 72 and 90 teeth respectively and comprises of m = 4 and 200 pressure angle gears. The motor is 8kW at 1000 rpm clockwise.

b. Calculate the rpm of the output shaft. c. Draw and calculate all the resultant

forces on all of the gears. d. Based on the above calculation,

discuss the relationship between torque and gear ratio.

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2

3

4

5

18T 18T

Figure 1

29

a) G3 G5

Comb 1 144 (8 x 18) 54 ( 3 x 18)

Comb 2 108 (6 x 18) 72 ( 4 x 18)

b)

05.0

90x72

18x18

NN

NNe

54

42

nL = e.nF

= (0.05)(1000)

= 50 rpm

c) FBD at Gear 2

Diameter pinion d = Nm

d = (18)(4x10-3)

= 0.072 m

Calculating the tangential for

kN122.2

N122,2

)1000)(072.0(

)000,8(60

dn

H60Ft

Direction of Ft: when gear 2 pushes gear 3, gear 3 will resist and therefore

the direction is UP

Fr= Ft tan 200 = 0.773kN

T2= Ft d/2 = (2,122)(0.072/2) = 76.4 Nm

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FBD at Gear 3

Reaction at gear 3 is equal in magnitude and opposite direction to gear 2

Diameter d3 = Nm

= (72) (4x10-3) m

= 0.288 m

T3 = Fr d3/2

= 2122 (0.288/2)

= 305.6 Nm

Rotation direction ccw

FBD at Gear 4

T3 must be equal to T4 with opposite direction. This is because the G3 and G4 is

on the same shaft, therefore, the total Torque = 0. Rotation direction ccw

T3 = T4 = 305.6 Nm

Gear 4 (ccw) pushes gear 3: dirction of Ft is DOWN.

Ft = 2 T/d

d = 0,072 m

Ft = 8.48 kN

Fr = 3.09 kN

Or

kN48.8

)250)(072.0(

)000,8(60

dn

H60Ft

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Gear 5